A certain circle can be represented by the following equation. $x^2+y^2+18x+14y+105=0$ What is the center of this circle ? $($
Explanation: The strategy We can find the center and radius of a circle by rewriting the given equation in the form of the standard equation of a circle. [What is the standard equation of the circle?] In order to do this, we take the following steps. Complete the square for both the $x^2$ and $y^2$ terms. [How do we complete the square?] Write the equation in the standard form of the circle. Completing the squares $\begin{aligned}x^2+y^2+18x+14y+105&=0\\\\ x^2+y^2+18x+14y&=-105\\\\ (x^2+18x)+(y^2+14y)&=-105 \text{(rearrange terms)}\\\\ (x^2+18x{+81})+(y^2+14y{+49})&=-105{+81}{+49}\end{aligned}$ Notice that we must add ${81}$ and ${49}$ on the right side of the equation, since we added them to the left side of the equation. [How did we get 81 and 49?] Writing the equation in standard form $\begin{aligned}(x^2+18x{+81})+(y^2+14y{+49})&=-105{+81}{+49}\\\\(x+9)^2+(y+7)^2&=25\\\\ (x-(-9))^2+(y-(-7))^2&=5^2\end{aligned}$ Since the equation is now in the standard form, we can conclude that this circle is centered at $(-9,-7)$ and has a radius of $5$ units. Summary The circle is centered at $(-9,-7)$. The circle has a radius of $5$ units.